# Quantum mechanics problem!

1. Jun 4, 2010

### QuantumJG

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

i.)

Normalisation implies that:

$$\int_{-\infty}^{\infty} \Psi (x,t)* \Psi (x,t) dx = 1$$

$$\int_{-\infty}^{\infty} |a_{0}| ^{2} | \psi _{0} (x) | ^ {2} + |a_{1}| ^{2} | \psi _{1} (x) | ^ {2} dx = 1$$

After a bit of work I got to this line

$$|a_{0}| ^{2} \int_{-\infty}^{\infty} | \psi _{0} (x) | ^ {2} dx + |a_{1}| ^{2} \int_{-\infty}^{\infty} | \psi _{1} (x) | ^ {2} dx = 1$$

$$|a_{0}| ^{2} + |a_{1}| ^{2} = 1$$

i.e. I found this to be the normalisation condition.

ii.)

First to see if the energy is an eigenvalue

$$[E] \Psi (x,t) = i \hbar \dfrac{d}{dt} \left( a _{0} \psi (x) e^{ - \dfrac{i E_{0} t }{ \hbar } } + a _{1} \psi (x) e^{ - \dfrac{i E_{1} t }{ \hbar } } \right)$$

$$[E] \Psi (x,t) = E_{0} a _{0} \psi (x) e^{ - \dfrac{i E_{0} t }{ \hbar } } + E_{1} a _{1} \psi (x) e^{ - \dfrac{i E_{1} t }{ \hbar } }$$

So the energy is not an eigenvalue and hence the energy is not a sharp observable.

$$\langle E \rangle = \int_{-\infty}^{\infty} \left( a _{0} * \psi (x) * e^{ \dfrac{i E_{0} t }{ \hbar } } + a _{1} * \psi (x) * e^{ \dfrac{i E_{1} t }{ \hbar } } \right) \left( E_{0} a _{0} \psi (x) e^{ - \dfrac{i E_{0} t }{ \hbar } } + E_{1} a _{1} \psi (x) e^{ - \dfrac{i E_{1} t }{ \hbar } } \right)$$

$$\langle E \rangle = E_{0} |a_{0}| ^{2} + E_{1} |a_{1}| ^{2}$$

$$\langle E \rangle = E_{0} |a_{0}| ^{2} + E_{1} (1 - |a_{0}| ^{2} )$$

$$\langle E \rangle = |a_{0}| ^{2} ( E_{0} - E_{1} ) + E_{1}$$

$$\langle E \rangle = |a_{0}| ^{2} ( \dfrac{1}{2} \hbar \omega - \dfrac{3}{2} \hbar \omega ) + \dfrac{3}{2} \hbar \omega$$

$$\langle E \rangle = \hbar \omega \left( \dfrac{3}{2} - |a_{0}| ^{2} \right)$$

I'm not 100% sure if this is right.

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2. Jun 4, 2010

### xepma

The mathematics is correct!

You might want to rephrase your answer to the question about the energy not being a sharp observable: saying "the energy is not an eigenvalue" does not really make sense.

PS it is also more standard to write H instead of |E|.

3. Jun 4, 2010

### QuantumJG

Ok but isn't a sharp observable one where if you apply it's corresponding operator to the wavefunction you get a constant times the wavefunction. The constant is an eigenvalue. But in this case the wavefunction isn't an eigenfunction, thus the observable isn't sharp.

It is more standard to apply the Hamiltonian operator, but the wavefunction isn't really in a form where if you apply the hamiltonian you will get something tangible.

4. Jun 5, 2010

### xepma

I was referring more to the way things were formulated. Saying "the energy is not an eigenvalue" does not really make sense. There is no "the energy"; the wavefunction is in a superposition. It's better to formulate it as: the wavefunction is not an eigenfunction of the Hamiltonian, and therefore energy is not a sharp observable.

Exactly in what way is tangible a criteria? The Hamiltonian is the operator associated with energy. But that does not imply that if you act on a wavefunction you obtain the corresponding energy.

5. Jun 10, 2010

### cartonn30gel

If you know about the matrix notation, it might make things easier to work out. That's how I would do this problem but your solution seems correct to me.